cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-3 of 3 results.

A085741 a(n) = T(n)^n, where T() are the triangular numbers (A000217).

Original entry on oeis.org

1, 1, 9, 216, 10000, 759375, 85766121, 13492928512, 2821109907456, 756680642578125, 253295162119140625, 103510234140112521216, 50714860157241037295616, 29345269354638035222576971
Offset: 0

Views

Author

Jon Perry, Jul 21 2003

Keywords

Examples

			a(3) = T(3)^3 = 6^3 = 216.
		

Crossrefs

Cf. A000217.
Essentially the same as A061718.

Programs

  • Magma
    [((n*(n+1))/2)^n: n in [0..20]]; // Vincenzo Librandi, Sep 14 2011
  • Maple
    a:=n->mul(sum(j, j=1..n),k=1..n): seq(a(n), n=0..13); # Zerinvary Lajos, Jun 02 2007
  • Mathematica
    With[{rnn=Range[20]},Join[{1},First[#]^Last[#]&/@Thread[ {Accumulate[ rnn],  rnn}]]] (* Harvey P. Dale, Dec 08 2013 *)
  • PARI
    a(n) = (n*(n+1)/2)^n; \\ Michel Marcus, Feb 19 2019
    

Formula

a(n) = ((n*(n+1))/2)^n. - Vincenzo Librandi, Sep 14 2011

Extensions

More terms from Ray Chandler, Nov 09 2003

A070307 Number of n X n matrices with nonnegative integer entries such that every row sum equals 3.

Original entry on oeis.org

1, 16, 1000, 160000, 52521875, 30840979456, 29509034655744, 42998169600000000, 90647430472564453125, 265599227914240000000000, 1047192117300356121695451136, 5410240907043328777415185924096, 35821862005173382840059779052734375, 298285661929377847941529600000000000000
Offset: 1

Views

Author

Sharon Sela (sharonsela(AT)hotmail.com), May 10 2002

Keywords

Crossrefs

Programs

  • Mathematica
    Table[ Binomial[n + 2, 3]^n, {n, 1, 14}]
  • PARI
    a(n) = binomial(n+2, 3)^n; \\ Michel Marcus, Mar 10 2017
    
  • Python
    import math
    f=math.factorial
    def C(n, r): return f(n)/ f(r)/ f(n-r)
    def A070307(n): return C(n + 2, 3)**n # Indranil Ghosh, Mar 10 2017

Formula

a(n) = C(n+2, 3)^n = A000292(n)^n.
a(n) ~ 6^(-n)*exp(3-5/(2*n))*n^(3*n). - Stefano Spezia, Jun 13 2025

Extensions

More terms from Robert G. Wilson v, May 10 2002
More terms from Michel Marcus, Mar 10 2017

A331988 Table T(n,k) read by antidiagonals. T(n,k) is the maximum value of Product_{i=1..n} Sum_{j=1..k} r_j[i] where each r_j is a permutation of {1..n}.

Original entry on oeis.org

1, 2, 2, 6, 9, 3, 24, 64, 20, 4, 120, 625, 216, 36, 5, 720, 7776, 3136, 512, 56, 6, 5040, 117649, 59049, 10000, 1000, 81, 7, 40320, 2097152, 1331000, 248832, 24336, 1728, 110, 8, 362880, 43046721, 35831808, 7529536, 759375, 50625, 2744, 144, 9, 3628800, 1000000000, 1097199376, 268435456, 28652616, 1889568, 93636, 4096, 182, 10
Offset: 1

Views

Author

Chai Wah Wu, Feb 23 2020

Keywords

Comments

A dual sequence to A260355. See arXiv link for sets of permutations that achieve the value of T(n,k). The minimum value of Product_{i=1..n} Sum_{j=1..k} r_j[i] is equal to n!*k^n.

Examples

			T(n,k)
   k    1    2     3      4      5      6      7      8      9     10     11     12
  ---------------------------------------------------------------------------------
n  1|   1    2     3      4      5      6      7      8      9     10     11     12
   2|   2    9    20     36     56     81    110    144    182    225    272    324
   3|   6   64   216    512   1000   1728   2744   4096   5832   8000  10648  13824
   4|  24  625  3136  10000  24336  50625  93636 160000 256036 390625 571536 810000
		

Crossrefs

Programs

  • Python
    from itertools import permutations, combinations_with_replacement
    def A331988(n,k): # compute T(n,k)
        if k == 1:
            count = 1
            for i in range(1,n):
                count *= i+1
            return count
        ntuple, count = tuple(range(1,n+1)), 0
        for s in combinations_with_replacement(permutations(ntuple,n),k-2):
            t = list(ntuple)
            for d in s:
                for i in range(n):
                    t[i] += d[i]
            t.sort()
            w = 1
            for i in range(n):
                w *= (n-i)+t[i]
            if w > count:
                count = w
        return count

Formula

T(n,n) = (n*(n+1)/2)^n = A061718(n).
T(n,k) <= (k(n+1)/2)^n.
T(1,k) = k = A000027(k).
T(n,1) = n! = A000142(n).
T(2,2m) = 9m^2 = A016766(m).
T(2,2m+1) = (3m+1)*(3m+2) = A001504(m).
T(n,2) = (n+1)^n = A000169(n+1).
T(3,k) = 8k^3 = A016743(k) for k > 1.
If n divides k then T(n,k) = (k*(n+1)/2)^n.
If k is even then T(n,k) = (k*(n+1)/2)^n.
If n is odd and k >= n-1 then T(n,k) = (k*(n+1)/2)^n.
If n is even and k is odd such that k >= n-1, then T(n,k) = ((k^2*(n+1)^2-1)/4)^(n/2).
Showing 1-3 of 3 results.